1
7.17.1 Formation of Ionic Bonds: Donating and AcceptingFormation of Ionic Bonds: Donating and Accepting
ElectronsElectrons
7.2 7.2 Energetics of Formation of Ionic CompoundsEnergetics of Formation of Ionic Compounds
7.3 7.3 Stoichiometry of Ionic CompoundsStoichiometry of Ionic Compounds
7.47.4 Ionic CrystalsIonic Crystals
7.57.5 Ionic RadiiIonic Radii
Ionic BondingIonic Bonding77
2
Bonding & Structure
3
Lewis Model G.N. Lewis in 1916
Only the outermost (valence) electrons are involved significantly in bond formation
Successful in solving chemical problems
4
Why are some elements so reactive (e.g.Na) and others inert (e.g. noble
gases)?
Why are there compounds with chemical formulae H2O and NaCl, but not H3O and NaCl2?
Why are helium and the other noble gases monatomic, when molecules of hydrogen and chlorine are diatomic?
5
Chemical bonds are strong electrostatic forces holding atoms or ions together, which are formed by the rearrangement (transfer or sharing) of outermost electrons
Atoms tend to form chemical bonds in such a way as to achieve the
electronic configurations of the nearest noble gases (The Octet Rule )
6
Ion s,p,d,f notationIsoelectronic
noble gas
Be2+
O2-
Sc3+
Br-
Ba2+
At-
Q.1 Write the s,p,d,f notation for the ions in the table below
7
Ion s,p,d,f notation Isoelectronic
noble gas
Be2+ 1s2 (2) He
O2- [He] 2s2,2p6 (2,8) Ne
Sc3+ [Ne] 3s2,3p6 (2,8,8) Ar
Br- [Ar] 3d10,4s2,4p6 (2,8,18,8) Kr
Ba2+ [Kr] 4d10,5s2,5p6 (2,8,18,18,8) Xe
At- [Xe] 4f14,5d10,6s2,6p6 (2,8,18,32,18,8) Rn
8
Three types of chemical bondsThree types of chemical bonds
1.Ionic bond (electrovalent bond)
Formed by transfer of electrons
Introduction (SB p.186)
9
Na Cl
Sodium atom, Na1s22s22p63s1
Chlorine atom, Cl1s22s22p63s23p5
7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)
Three types of chemical bondsThree types of chemical bonds
1.Ionic bond (electrovalent bond)
10
Three types of chemical bondsThree types of chemical bonds
1.Ionic bond
Introduction (SB p.186)
Electrostatic attraction between positively charged particles and negatively charged particles
11
Three types of chemical bondsThree types of chemical bonds
2. Covalent bond
Formed by sharing of electrons
Introduction (SB p.186)
12
Three types of chemical bondsThree types of chemical bonds
2.Covalent bond
Introduction (SB p.186)
Electrostatic attraction between nuclei and shared electrons
13
Three types of chemical bondsThree types of chemical bonds
3.Metallic bond
Electrostatic attraction between metallic cations and delocalized electrons (electrons that have no fixed positions)
Introduction (SB p.186)
14
Three types of chemical bondsThree types of chemical bonds
3. Metallic bond
Introduction (SB p.186)
Formed by sharing of a large number of delocalized electrons
15
Ionic bonds and Covalent bonds are only extreme cases of a continuum.
In real situation, most chemical bonds are intermediate between ionic and covalent
16
Ionic Bonds with incomplete transfer of electrons have covalent character.Covalent Bonds with unequal sharing of electrons have ionic character.
17
Electronegativity and Types of Chemical BondsIonic or covalent depends on the electron-attracting ability of bonding atoms in a chemical bond.
Ionic bonds are formed between atoms with great difference in their electron-attracting abilities
Covalent bonds are formed between atoms with small or no difference in their electron-attracting abilities.
18
Ways to compare the electron-attracting ability of atoms
1. Ionization Enthalpy
2. Electron affinity
3. Electronegativity
19
Electronegativity and Types of Chemical Bonds1. Ionization enthalpy
The enthalpy change when one mole of electrons are removed from one mole of atoms or positive ions in gaseous state.
X(g) X+(g) + e- H 1st
I.E.
X+(g) X2+(g) + e- H 2nd
I.E.
Ionization enthalpies are always positive.
20
X(g) + e- X-(g) H 1st
E.A.
X(g) + e- X2(g) H 2nd
E.A.
Electronegativity and Types of Chemical Bonds2. Electron affinity
The enthalpy change when one mole of electrons are added to one mole of atoms or negative ions in gaseous state.
Electron affinities can be positive or negative.
21
I.E. and E.A. only show e- releasing/attracting power of
free, isolated atoms
However, whether a bond is ionic or covalent depends on the ability of atoms to attract electrons in a chemical bond.
22
Electronegativity and Types of Chemical Bonds3. Electronegativity
The ability of an atom to attract electrons in a chemical bond.
23
Mulliken’s scale of electronegativity
Electronegativity(Mulliken) )(2
1EAIE
Nobel Laureate in Chemistry, 1966
24
Pauling’s scale of electronegativity
Nobel Laureate in Chemistry, 1954
Nobel Laureate in Peace, 1962
25
Pauling’s scale of electronegativity calculated from bond enthalpies
cannot be measured directly
having no unit
always non-zero
the most electronegative element, F, is arbitrarily assigned a value of 4.00
26
I A I I A I I I A IVA VA VIA VI IA VI I IA
H 2.1
He -
Li 1.0
Be 1.5
B 2.0
C 2.5
N 3.0
O 3.5
F 4.0
Ne -
Na 0.9
Mg 1.2
Al 1.5
Si 1.8
P 2.1
S 2.5
Cl 3.0
Ar -
K 0.8
Ca 1.0
Ga 1.6
Ge 1.8
As 2.0
Se 2.4
Br 2.8
Kr -
Rb 0.8
Sr 1.0
I n 1.7
Sn 1.8
Sb 2.0
Te 2.1
I 2.5
Xe -
Cs 0.7
Ba 0.9
Tl 1.8
Pb 1.8
Bi 1.9
Po 2.0
At 2.2
Rn -
Pauling’s scale of electronegativity
27
What trends do you notice about the EN values in the Periodic Table? Explain. I A I I A I I I A IVA VA VIA VI IA VI I IA
H 2.1
He -
Li 1.0
Be 1.5
B 2.0
C 2.5
N 3.0
O 3.5
F 4.0
Ne -
Na 0.9
Mg 1.2
Al 1.5
Si 1.8
P 2.1
S 2.5
Cl 3.0
Ar -
K 0.8
Ca 1.0
Ga 1.6
Ge 1.8
As 2.0
Se 2.4
Br 2.8
Kr -
Rb 0.8
Sr 1.0
I n 1.7
Sn 1.8
Sb 2.0
Te 2.1
I 2.5
Xe -
Cs 0.7
Ba 0.9
Tl 1.8
Pb 1.8
Bi 1.9
Po 2.0
At 2.2
Rn -
28
I A I I A I I I A IVA VA VIA VI IA VI I IA
H 2.1
He -
Li 1.0
Be 1.5
B 2.0
C 2.5
N 3.0
O 3.5
F 4.0
Ne -
Na 0.9
Mg 1.2
Al 1.5
Si 1.8
P 2.1
S 2.5
Cl 3.0
Ar -
K 0.8
Ca 1.0
Ga 1.6
Ge 1.8
As 2.0
Se 2.4
Br 2.8
Kr -
Rb 0.8
Sr 1.0
I n 1.7
Sn 1.8
Sb 2.0
Te 2.1
I 2.5
Xe -
Cs 0.7
Ba 0.9
Tl 1.8
Pb 1.8
Bi 1.9
Po 2.0
At 2.2
Rn -
The EN values increase from left to right across a Period.
29
What trends do you notice about the EN values in the Periodic Table? Explain.The EN values increase from left to right across a Period.
The atomic radius decreases from left to right across a Period.Thus,the nuclear attraction experienced by the bonding electrons increases accordingly.
30
The EN values decrease down a Group. I A I I A I I I A IVA VA VIA VI IA VI I IA
H 2.1
He -
Li 1.0
Be 1.5
B 2.0
C 2.5
N 3.0
O 3.5
F 4.0
Ne -
Na 0.9
Mg 1.2
Al 1.5
Si 1.8
P 2.1
S 2.5
Cl 3.0
Ar -
K 0.8
Ca 1.0
Ga 1.6
Ge 1.8
As 2.0
Se 2.4
Br 2.8
Kr -
Rb 0.8
Sr 1.0
I n 1.7
Sn 1.8
Sb 2.0
Te 2.1
I 2.5
Xe -
Cs 0.7
Ba 0.9
Tl 1.8
Pb 1.8
Bi 1.9
Po 2.0
At 2.2
Rn -
31
The EN values decrease down a Group.The atomic radius increases down a Group, thus weakening the forces of attraction between the nucleus and the bonding electrons.
What trends do you notice about the EN values in the Periodic Table? Explain.
32
I A I I A I I I A IVA VA VIA VI IA VI I IA
H 2.1
He -
Li 1.0
Be 1.5
B 2.0
C 2.5
N 3.0
O 3.5
F 4.0
Ne -
Na 0.9
Mg 1.2
Al 1.5
Si 1.8
P 2.1
S 2.5
Cl 3.0
Ar -
K 0.8
Ca 1.0
Ga 1.6
Ge 1.8
As 2.0
Se 2.4
Br 2.8
Kr -
Rb 0.8
Sr 1.0
I n 1.7
Sn 1.8
Sb 2.0
Te 2.1
I 2.5
Xe -
Cs 0.7
Ba 0.9
Tl 1.8
Pb 1.8
Bi 1.9
Po 2.0
At 2.2
Rn -
Why are the E.A. of noble gases not shown ?
33
EN is a measure of the ability of an atom to attract electrons in a chemical bond.
However, noble gases are so inert that they rarely form chemical bonds with other atoms.
Why are the E.A of noble gases not shown ?
XeF2, XeF4 and XeF6 are present
34
Formation of Ionic BondAtoms of Group IA and IIA elements ‘tend’ to achieve the noble gas structures in the previous Period by losing outermost electron(s).
In fact, formations of positive ions from metals are endothermic and not spontaneous.
I.E. values are always positive
35
Formation of Ionic BondAtoms of Group VIA and VIIA elements tend to achieve the noble gas structures in the same Periods by gaining electron(s)
First E.A. of Group VIA and VIIA elements are always negative.
Spontaneous and exothermic processes
36
The oppositely charged ions are stabilized by coming close to each other to form the ionic bond.
Ionic bond is the result of electrostatic interaction between oppositely charged ions.
Interaction = attraction + repulsion
37
Dots and Crosses Diagram
38
Notes on Dots & Crosses representation Electrons in different atoms are
indistinguishable. The dots and crosses do not indicate
the exact positions of electrons. Not all stable ions have the noble
gas structures(Refer to pp.4-5).
39
Tendency for the Formation of IonsAn ion will be formed easily if 1. The electronic structure of the ion
is stable;2. The charge on the ion is small;3. The size of parent atom from which
the ion is formed is small for an anion, or
large for a cation.
40
For cation,
Larger size of parent atom
less positive I.E. or less +ve sum of successive I.E.s, easier formation of cation,
atoms of Group IA & Group IIA elements form cations easily.
41
Li+ Be2+
Na+ Mg2+ Al3+
K+ Ca2+ Sc3+
Rb+ Sr2+ Y3+ Zr4+
Cs+ Ba2+ La3+ Ce4+
Ease of formation of cation Ease of formation of cation
42
For anion,Smaller size of parent atom
more negative E.A. or more -ve sum of
successive E.A.s,
easier formation of anion, atoms of Group VIA & Group VIIA elements form anions easily.
43
N3 O2 F
P3 S2 Cl
Br
I
Ease of formation of anion
Ease
of fo
rmatio
n o
f an
ion
44
Stable Ionic StructuresNot all stable ions have the noble gas electronic structures.
Q.2 Write the s,p,d,f notation for each of the following cations.
Use your answers to identify three types of commonly occurring arrangement of outershell electrons of cations other than the stable octet structure.
45
Ion s,p,d,f notationZn2+
Pb2+
V3+
Cr3+
Fe3+
Fe2+
Co2+
Ni2+
Cu2+
Q.2 Write the s,p,d,f notation for the following cations
46
Ion s,p,d,f notationZn2+ 1s2,2s2,2p6,3s2,3p6,3d10
Pb2+ 1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2
V3+ 1s2,2s2,2p6,3s23p63d2
Cr3+ 1s2,2s2,2p6,3s23p63d3
Fe3+ 1s2,2s2,2p6,3s23p63d5
Fe2+ 1s2,2s2,2p6,3s23p63d6
Co2+ 1s2,2s2,2p6,3s23p63d7
Ni2+ 1s2,2s2,2p6,3s23p63d8
Cu2+ 1s2,2s2,2p6,3s23p63d9
Q.2
47
Ion s,p,d,f notationZn2+ 1s2,2s2,2p6,3s2,3p6,3d10
Pb2+ 1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2
V3+ 1s2,2s2,2p6,3s23p63d2
Cr3+ 1s2,2s2,2p6,3s23p63d3
Fe3+ 1s2,2s2,2p6,3s23p63d5
Fe2+ 1s2,2s2,2p6,3s23p63d6
Co2+ 1s2,2s2,2p6,3s23p63d7
Ni2+ 1s2,2s2,2p6,3s23p63d8
Cu2+ 1s2,2s2,2p6,3s23p63d9
1. 18 - electrons group e.g. Zn2+
Fully-filled s, p & d-subshells
48
Ion s,p,d,f notationZn2+ 1s2,2s2,2p6,3s2,3p6,3d10
Pb2+ 1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2
V3+ 1s2,2s2,2p6,3s23p63d2
Cr3+ 1s2,2s2,2p6,3s23p63d3
Fe3+ 1s2,2s2,2p6,3s23p63d5
Fe2+ 1s2,2s2,2p6,3s23p63d6
Co2+ 1s2,2s2,2p6,3s23p63d7
Ni2+ 1s2,2s2,2p6,3s23p63d8
Cu2+ 1s2,2s2,2p6,3s23p63d9
Fully-filled s, p & d-subshells
2. (18 + 2 ) - electron group e.g. Pb2+
49
Ion s,p,d,f notationZn2+ 1s2,2s2,2p6,3s2,3p6,3d10
Pb2+ 1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2
V3+ 1s2,2s2,2p6,3s23p63d2
Cr3+ 1s2,2s2,2p6,3s23p63d3
Fe3+ 1s2,2s2,2p6,3s23p63d5
Fe2+ 1s2,2s2,2p6,3s23p63d6
Co2+ 1s2,2s2,2p6,3s23p63d7
Ni2+ 1s2,2s2,2p6,3s23p63d8
Cu2+ 1s2,2s2,2p6,3s23p63d9
3. Variable arrangements for ions of transition metals. ns2,np6, nd1 to ns2,np6, nd9
Ti3+ [Ne] 3s23p63d1 Mn3+ [Ne] 3s23p63d4
50
Electronic arrangements of stable cations
Ionization enthalpy determines the ease of formation of cations.
Less positive I.E. or sum of I.E.s
easier formation of cations
51
Electronic arrangements of stable cationsOctet structure : -cations of Group IA, IIA and IIIB
I A I IA I I I B
2,8 Na+ Mg2+ Al3+(I I IA)
2,8,8 K+ Ca2+ Sc3+
2,8,18,8 Rb+ Sr2+ Y3+
52
(a) 18 - electrons group
cations of Groups IB, IIB, IIIA and IVA
I B I IB I I IA IVA
2,8,18 Cu+ Zn2+ Ga3+ - -
2,8,18,18 Ag+ Cd2+ I n3+ Sn4+
2,8,18,32,18 Au+ Hg2+ Tl3+ Pb4+
53
I B I IB I I IA IVA
2,8,18 Cu+ Zn2+ Ga3+ - -
2,8,18,18 Ag+ Cd2+ I n3+ Sn4+
2,8,18,32,18 Au+ Hg2+ Tl3+ Pb4+
Less stable than ions with a noble gas structure.
54
I B I IB I I IA IVA
2,8,18 Cu+ Zn2+ Ga3+ - -
2,8,18,18 Ag+ Cd2+ I n3+ Sn4+
2,8,18,32,18 Au+ Hg2+ Tl3+ Pb4+
Cu and Au form other ions because the nuclear charges are not high enough to hold the 18-electron group firmly.
55
I B I IB I I IA IVA
2,8,18 Cu+ Zn2+ Ga3+ - -
2,8,18,18 Ag+ Cd2+ I n3+ Sn4+
2,8,18,32,18 Au+ Hg2+ Tl3+ Pb4+
Cu2+ 2, 8, 17 Au3+ 2, 8, 18, 32, 16The d-electrons are more diffused and thus less attracted by the less positive nuclei.
56
(b) (18+2) electrons
Cations of heavier group members
due to presence of 4f electrons.
Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1
Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)
Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)
Stability : Tl+ > Tl3+ due to extra stability of
6s electrons (inert pair effect)
57
(b) (18+2) electrons
6s electrons are poorly shielded by the inner 4f electrons.
6s electrons experience stronger nuclear attraction.
Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1
Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)
Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)
58
(b) (18+2) electrons
Inert pair effect increases down a Group.
Stability :Sn4+(18) > Sn2+(18+2)
Pb2+(18+2) > Pb4+(18)
moving down Group IV
Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1
Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)
Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)
59
(b) (18+2) electrons
Inert pair effect increases down a Group.
Stability :Sb5+(18) > Sb3+(18+2)
Bi3+(18+2) > Bi5+(18)
moving down Group V
Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1
Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)
Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)
60
(c) Cations of transition elements
- variable oxidation numbers
- Electronic configurations from
ns2, np6, nd1 to ns2, np6, nd9
Fe [Ne] 3s2, 3p6, 3d6, 4s2
Fe2+ [Ne] 3s2, 3p6, 3d6
Fe3+ [Ne] 3s2, 3p6, 3d5
61
(c) Cations of transition elements
Which one is more stable, Fe2+(g) or Fe3+(g) ?
Fe2+(g) is more stable than Fe3+(g)
Energy is needed to remove electrons from Fe2+(g) to give Fe3+(g)
Fe [Ne] 3s2, 3p6, 3d6, 4s2
Fe2+ [Ne] 3s2, 3p6, 3d6
Fe3+ [Ne] 3s2, 3p6, 3d5
62
B. Anions - with noble gas structures
Electron affinity determines the ease of formation of anions.
More -ve E.A. or sum of E.A.s
more stable anion
Group VIA and Group VIIA elements form anions easily.
63
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-322
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-325
Kr+39
First Electron Affinity (kJ mol1)
X(g) + e X(g)
64
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-328
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-324
Kr+39
E.A. becomes more –ve from Gp 1 to Gp 7 across a period
65
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-328
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-324
Kr+39
The electrons added experience stronger nuclear attraction when the atoms are getting smaller across the period.
66
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-328
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-324
Kr+39
+ve E.A. for Gp 2 and Gp 0 because the electron added occupies a new shell / subshell
67
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-328
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-324
Kr+39
Goes to a new subshell Goes to a new shellBe(2s2) Be(2s22p1)
Ne(2p6) Ne(2p63s1)
68
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-328
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-324
Kr+39
More –ve E.A. for Gp 1 because the ions formed have full-filled s-subshells.
E.g. Li(1s22s1) Li(1s22s2)
69
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-328
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-324
Kr+39
Less –ve E.A. for Gp 5 because the stable half-filled p-subshell no longer exists.
E.g. N(2s22p3) N(2s22p4)
70
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-328
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-324
Kr+39
Q.3 Why are the first E.A.s of halogen atoms more negative than those of the O atom or the S atom ?
71
H-73
He+21
Li-60
Be+18
B-23
C-122
N~0
O-142
F-328
Ne+29
Na-53
Mg+21
Al-44
Si-134
P-72
S-200
Cl-349
Ar+35
K-48
Br-324
Kr+39
It is because the halide ions formed have full-filled s/p subshells (octet structure).
72
Q.4 Why are the second E.A.s of O(+844 kJmol1) and S(+532 kJmol1) positive ?
The electrons added are repelled strongly by the negative ions.
O(g) + e O2(g)
S(g) + e S2(g)
73
Why is the E.A. of F less negative than that of Cl ?
The size of Fluorine atom is so small that the addition of an extra electron results in great repulsion among the electrons.
Halogen
F Cl Br I
E.A.kJ mol1
328 349 325 295
The 2nd electron shell is much smaller than the 3rd electron shell.
74
Energetics of Formation of Ionic CompoundsA. Formation of an ion pair in gaseous state
Consider the formation of a KF(g) ion pair from K(g) & F(g)
H1K(g) + F(g) KF(g)
IE1(K)
K+(g)
H2
By Hess’s law, H1 = IE1(K) + EA1(F) + H2
EA1(F)
+ F(g)
75
R4π
mol106.022
mol J 10328.0
mol106.022
mol J 10419.0
0
21123
13
123
13
m102.170mJC 108.8544πC)10(1.602
J101.511 1011212
21919
J19109.119
H1 = IE1(K) + EA1(F) + H2
Q.6
J101.063J101.511 1819
(+)()
76
J109.119ΔH 191
H1 = IE1(K) + EA1(F) + H2
= 1.5111019 J 1.0631018 J = 9.1191019 J
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
77
When R
Stability : K+(g) + F-(g) < K(g) + F(g) by +1.511x10-19J
J109.119ΔH 191
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
78
J109.119ΔH 191
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
I.E. and E.A. are NOT the driving forces for the formation of ionic bond.
79
J109.119ΔH 191
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
K+(g) & F(g) tend to come close together in order to become stable.
80
J109.119ΔH 191
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
Coulomb stabilization is the driving force for the formation of ionic bond.
81
J109.119ΔH 191
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
When R = 2.1701010 m
The most energetically stable state is reached.
82
J109.119ΔH 191
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
The ions cannot come any closer than Re as it will results in less stable states
83
J109.119ΔH 191
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
Repulsions between electron clouds and between nuclei > attraction between ions
84
R
Pote
ntia
l en
erg
y
(V)
Minimum V when R = 2.170 1010 m
Repulsion :
12R
1V
between opposite charges
Attraction :R
1V
between opposite charges
85
J109.119ΔH 191
IE1(K) + EA1(F) = +1.5111019 J
H2 = 1.0631018 J
What is the significance of the lowest energy state of the neutral state of K + F ?
86
K F
Electrostatic attraction between positive nuclei and bond electron pair stabilizes the system
K FA covalent bond is formed
87
A. Formation of Ionic Crystals
Consider the formation of NaCl(s) from Na(s) & Cl2(g)
H2
By Hess’s law, Hf = H1 + H2
Na+(g) + Cl(g)
H1
HfNa(s) + Cl2(g) NaCl(s) 21
88
H1 is the sum of four terms of enthalpy changes
1. Standard enthalpy change of atomization of Na(s)
Na(s) Na(g) H = +108.3 kJ mol-1
2. First ionization enthalpy of Na(g)
Na(g) Na+(g) + e- H= +500 kJ mol-1
3. Standard enthalpy change of atomization of Cl2(g)
1/2Cl2(g) Cl(g) H = +121.1 kJ mol-1
4. First electron affinity of Cl(g)
Cl(g) + e- Cl-(g) H = -349 kJ mol-1
89
H2 is the lattice enthalpy of NaCl.
It is the enthalpy change for the formation of 1 mole of NaCl(s) from its constituent ions in the gaseous state.
Na+(g) + Cl-(g) NaCl(s) HLo
90
1.theoretical calculation using an ionic model, Or
2.experimental results indirectly with the use of a Born-Haber cycle.
Direct determination of lattice enthalpy by experiment is very difficult, but it can be obtained from
91
Q.7 Calculate the lattice enthalpy of NaClHf = H1 +H2
Lattice enthalpy
= H2
= Hf - H1
= [(411) (+108.3 + 500 + 121.1 349)] kJ mol1
= 791.4 kJ mol1
92
-349
-791.4
93
Lattice enthalpy is the dominant enthalpy term responsible for the -ve Hf of an ionic compound.
More ve HLo More stable ionic compound
Lattice enthalpy is a measure of the strength of ionic bond.
94
Determination of Lattice Determination of Lattice EnthalpyEnthalpy
• From Born-Haber cycle (experimental method, refers to Q.7)
• From theoretical calculation based on the ionic model
95
Assumptions made in the Assumptions made in the calculationcalculation
• Ions are spherical and have no distortion of the charge cloud, I.e. 100% ionic.
• The crystal has certain assumed lattice structure.
• Repulsive forces between oppositely charged ions at short distances are ignored.
96
)(4 0
rr
QMLQH lattice
M is the Madelung constants that depends on the crystal structure
97
)(4 0
rr
QMLQH lattice
L is the Avogadro’s constant
98
)(4 0
rr
QMLQH lattice
Q+ and Q- are the charges on the cation and the anion respectively
99
)(4 0
rr
QMLQH lattice
0 is the permittivity of vacuum
100
)(4 0
rr
QMLQH lattice
(r+ + r-) is the nearest distance between the nuclei of cation and anionr+ is the ionic radius of the cation r- is the ionic radius of the anion
101
Stoichiometry of Ionic CompoundsStoichiometry of Ionic Compounds
Stoichiometry of an ionic compound is the simplest whole number ratio of cations and anions involved in the formation of the compound. There are two ways to predict stoichiometry.
102
1. Considerations in terms of electronic configurations
Atoms tend to attain noble gas electronic structures by losing or gaining electron(s).
The ionic compound is electrically neutral.
total charges on cation = total charges on anionstotal charges on cation = total charges on anions
103
1. Considerations in terms of electronic configurations
Na + F [Na]+ [F]-
2,8,1 2,7 2,8 2,8
Mg + 2Cl [Cl]- [Mg]2+ [Cl]-
2,8,2 2,8,7 2,8,8 2,8 2,8,8
NaF
MgCl2
104
2. Considerations in terms of enthalpy changes of formation
Based on the Born-Haber cycle & the ththeoretically calculated lattice enthalpyeoretically calculated lattice enthalpy, the values of Hf
o of hypothetical compounds (e.g. MgCl , MgCl2 , MgCl3) can be calculated.
The stoichiometry with the most negative The stoichiometry with the most negative HHff
o o value is the most stable onevalue is the most stable one.
105
2. Considerations in terms of enthalpy changes of formation
Or, we can determine the true stoichiometry by comparing the calculated Hf
o values of hypothetical compounds with the experimentally determined Hf
o value.
The stoichiometry with Hfo value closest to th
e experimentally determined Hfo value is the ans
wer.
106
Q.8 Three hypothetical formulae of magnesium chloride are proposed and their estimated lattice enthalpies are shown in the table below.
Hypothetical stoichiometry
MgCl MgCl2 MgCl3
Assumed structure
NaCl CaF2 AlCl3
Estimated HL
o(kJ mol- 1) - 771 - 2602 - 5440
107
Hf[MgCl(s)] = Hat[Mg(s)] + 1st IE of Mg + Hat[1/2Cl2(g)]
+ 1st EA of Cl + HL[MgCl(s)]
= (+150 + 736 + 121 - 364- 771) kJ mol-1
= -128 kJ mol-1
108
Hat[1/2Cl2]
Mg+(g)+Cl(g)
Mg+(g)+Cl(g)1st EA[Cl]
MgCl(s)
HL[MgCl]
Hf[MgCl]<0
1st IE[Mg]
Mg+(g)+ Cl2(g)21
Enthalpy (kJ m
ol
)
Mg(s)+ Cl2(g)21
Hat[Mg]
Mg(g)+ Cl2(g)21
109
Hf[MgCl2(s)]
= Hat[Mg(s)] + 1st IE of Mg + 2nd IE of Mg
+ 2Hat[1/2Cl2(g)] + 21st EA of Cl + HL[MgCl2(s)]
= [150+736+1450+2121+2(-364)+(-2602)] kJ mol-1
= -752 kJ mol-1
110
Enthalpy (kJ
mol
)
Mg(s)+Cl2(g)
Mg(g)+Cl2(g)
Hat[Mg]
Mg+(g)+Cl2(g)
1st IE[Mg]
Mg2+(g)+Cl2(g)
2nd IE[Mg]
Mg2+(g)+2Cl(g)
2Hat[1/2Cl2]
Mg2+(g)+2Cl(g)
21st EA[Cl]
MgCl2(s)
HL[MgCl2]
Hf[MgCl2] <0
111
Hf[MgCl3(s)]
= Hat[Mg(s)] + 1st IE of Mg + 2nd IE of Mg + 3rd IE of Mg
+ 3Hat[1/2Cl2(g)] + 31st EA of Cl + HL[MgCl3(s)]
= 150+736+1450+7740+3121+3(-364)+(-5440)
= +3907 kJ mol-1
112
Enthalpy ( kJ m
ol
)
Mg(s)+3/2Cl2(g)
Mg3+(g)+3Cl(g)31st EA[Cl]
MgCl3(s)
HL[MgCl3]
Hf[MgCl3] >0
Mg3+(g)+3Cl(g)
3Hat[1/2Cl2]
3rd IE[Mg]
Mg3+(g)+ Cl2(g)23
Mg2+(g)+ Cl2(g)23
2nd IE[Mg]Mg+(g)+ Cl2(g)
23
1st IE[Mg]
Hat[Mg]
Mg(g)+ Cl2(g)23
113
Since the hypothetical compound MgCl2 has the most negative Hf value, and
this value is closest to the experimentally determined one,
the most probable formula of magnesium chloride is
MgCl2
114
Reasons for the discrepancy between calculated and experimental results of Hf
The ionic bond of MgCl2 is not 100% pure. I,e. the ions are not perfectly spherical.
MgCl2 has a different crystal structure from CaF2
Short range repulsive forces between oppositely charged ions have not been considered.
Check Point 7-2Check Point 7-2
115
7.2 Energetics of Formation of Ionic Compounds (SB p.196)
Factors affecting lattice enthalpyFactors affecting lattice enthalpy
• Effect of ionic size:
The greater the ionic size The lower (or less negative) is the lattice enthalpy
)(4 0
rr
QMLQH lattice
116
7.2 Energetics of Formation of Ionic Compounds (SB p.196)
Factors affect lattice enthalpyFactors affect lattice enthalpy
• Effect of ionic charge:
The greater the ionic charge The higher (or more negative) is the
lattice enthalpy
Check Point 7-3Check Point 7-3
)(4 0
rr
QMLQH lattice
117
7.7.44Ionic CrystalsIonic Crystals
118
7.4 Ionic Crystals (SB p.202)
Crystal LatticeCrystal Lattice(( 晶體格子晶體格子 ))
119
7.4 Ionic Crystals (SB p.202)
In solid state, ions of ionic compounds are In solid state, ions of ionic compounds are regularly packedregularly packed to form 3-dimensional str to form 3-dimensional structures called uctures called crystal latticescrystal lattices(( 晶體格子晶體格子 ))
120
7.4 Ionic Crystals (SB p.202)
The The coordination numbercoordination number (C.N.) of a given partic (C.N.) of a given particle in a crystal lattice is the le in a crystal lattice is the number of nearest neinumber of nearest neighbours of the particleghbours of the particle..
C.N. of Na+
= 6
121
7.4 Ionic Crystals (SB p.202)
The The coordination numbercoordination number (C.N.) of a given partic (C.N.) of a given particle in a crystal lattice is the le in a crystal lattice is the number of nearest neinumber of nearest neighbours of the particleghbours of the particle..
C.N. of Cl
= 6
122
Identify the unit cell of NaCl crystal latticeIdentify the unit cell of NaCl crystal lattice
Not a unit Not a unit cellcell
123
Identify the unit cell of NaCl crystal latticeIdentify the unit cell of NaCl crystal lattice
Not a unit Not a unit cellcell
124
The The unit cellunit cell of a crystal lattice is the simplest of a crystal lattice is the simplest 3-D arrangement of particles which, when 3-D arrangement of particles which, when repeated 3-dimensionallyrepeated 3-dimensionally in space, will in space, will generate the whole crystal lattice.generate the whole crystal lattice.
125
The The unit cellunit cell of a crystal lattice is the simplest of a crystal lattice is the simplest 3-D arrangement of particles which, when 3-D arrangement of particles which, when repeated 3-dimensionally in space, will repeated 3-dimensionally in space, will generate the whole crystal lattice.generate the whole crystal lattice.
126
The The unit cellunit cell of a crystal lattice is the simplest of a crystal lattice is the simplest 3-D arrangement of particles which, when 3-D arrangement of particles which, when repeated 3-dimensionally in space, will repeated 3-dimensionally in space, will generate the whole crystal lattice.generate the whole crystal lattice.
127
Unit Unit cellcell
Unit Unit cell cell
128
A crystal lattice with a A crystal lattice with a cubic unit cubic unit cellcell is known as a is known as a cubic structurecubic structure..
129
Three Kinds of Cubic Three Kinds of Cubic StructureStructure• Simple cubic (primitive) structure• Body-centred cubic (b.c.c.) structure• Face-centred cubic (f.c.c.) structure
130
Simple cubic structure
View this Chemscape 3D structure
Unit cell
Space filling
Space lattice
131
Body-centred cubic (b.c.c.) structureBody-centred cubic (b.c.c.) structure
View this Chemscape 3D structure
Unit cell
Space filling
Space lattice
132
Face-centred cubic (f.c.c.) structureFace-centred cubic (f.c.c.) structure
View this Chemscape 3D structure
Unit cell
133
The f.c.c. structure can be obtained by stacking the layers of particles in the pattern abcabc...
134
Interstitial sitesInterstitial sites
- The empty spaces in a crystal - The empty spaces in a crystal latticelatticeTwo types of interstitial sites in f.c.c. structureOctahedral siteTetrahedral site
135
Octahedral site : -
It is the space between the 3 spheres in one layer and 3 other spheres in the adjacent layer.
136
When the octahedron is rotated by 45o , the octahedral site can also be viewed as the space confined by 4 spheres in one layer and 1 other sphere each in the upper and lower layers respectively.
45o
beforebehind
137
Tetrahedral site : -It is the space between the 3 spheres of one layer and a fourth sphere on the upper layer.
138
In an f.c.c. unit cell, the tetrahedral site is the space bounded by a corner atom and the three face-centred atoms nearest to it.
139
c
b a
a
140
Q.9 Label a, b, and c as Oh sites or Td sites
a and b are Td sites
c is Oh site
Top layer
141
Q.10 Identify the Oh sites and Td sites in the f.c.c. unit cell shown below.
13 Oh sites
3
5 6
89
1011
1213
4 Th sites in the front
1
2
35 6
7 8
12
4
4
7 4 Th sites at the back
View Octahedral sites and tetrahedral sites
142
Since ionic crystal lattice is made of two kinds of particles, cations and anions,
the crystal structure of an ionic compound can be considered as two lattices of cations and anions interpenetrating with each other.
143
The crystal structures of
sodium chloride,
caesium chloride and
calcium fluoride are discussed.
144
Sodium Chloride (The Rock Salt Structure)
f.c.c. unit cell of larger Cl- ions, with all Oh sites occupied by the smaller Na+ ions
Oh sites of f.c.c. lattice of Cl- ions
145
A more open f.c.c. unit cell of smaller Na+ ions with all Oh sites occupied by the larger Cl- ions.
Oh sites of f.c.c. lattice of Na+ ions
146
F.C.C. Structure of Sodium Chloride
View this Chemscape 3D structure
147
Co-ordination number of Na+ = 6Co-ordination number of Cl- = 6
6 : 6 co-ordination
Unit cell of NaClStructure of Sodium ChlorideStructure of Sodium Chloride
7.4 Ionic Crystals (SB p.201)
148
Only the particles in the centre (or body) of the unit cell belong to the unit cell entirely. Particles locating on the faces, along the edges or at the corner of a unit cell are shared with the neighboring unit cells of the crystal lattice.
149
Fraction of particles occupied by a unit cell
1/21
CornerEdgeFaceBody
1/2
150
Fraction of particles occupied by a unit cell
1/41/21
CornerEdgeFaceBody
1/4
151
Fraction of particles occupied by a unit cell
1/81/41/21
CornerEdgeFaceBody
1/8
152
Q.11 Calculate the net nos. of Na+ and Cl- ions in a unit cell of NaCl.
No. of Cl- ions = 42
16
8
18
No. of Na+ ions = 414
112
8 corners 6 faces
12 edges
1 body
Example 7-4Example 7-4
153
Structure of Caesium Chloride Structure of Caesium Chloride Link Link
7.4 Ionic Crystals (SB p.202)
Simple cubic latticeCs+ ions are too large to fit in the octahedral sites.
Thus Cl ions adopt the more open simple cubic structure with the cubical sites occupied by Cs+ ions.
154
A cubical site is the space confined by 4 spheres in one layer and 4 other spheres in the adjacent layer.
155
Size of interstitial sites : -
Cubical > octahedral > tetrahedral
In f.c.c. structure
In simple cubic structure
156
Co-ordination number of Cs+ = 8
Co-ordination number of Cl- = 88 : 8 co-ordination
7.4 Ionic Crystals (SB p.202)
Simple cubic lattice
Structure of Caesium Chloride Structure of Caesium Chloride Link Link
157
Q.12
Number of Cs+ = 1 1
8
18 Number of Cl =
158
Structure of Calcium Fluoride Structure of Calcium Fluoride Link Link
7.4 Ionic Crystals (SB p.203)
It can be viewed as a simple cubic structure of larger fluoride ions with alternate cubical sites occupied by smaller calcium ions.
Fluorite structure
159
7.4 Ionic Crystals (SB p.203)
Alternately, it can be viewed as an expanded f.c.c. structure of smaller calcium ions with all tetrahedral sites occupied by larger fluoride ions.
Structure of Calcium Fluoride Structure of Calcium Fluoride Link Link
160
Co-ordination number of Ca2+ = 8
Co-ordination number of F- = 48 : 4 co-ordination
7.4 Ionic Crystals (SB p.203)
Face-centred cubic lattice
Structure of Calcium Fluoride Structure of Calcium Fluoride Link Link
161
7.4 Ionic Crystals (SB p.203)
Q.13 (a)
Number of F = 8
Number of Ca2+ = 42
16
8
18
CaF2
162
7.4 Ionic Crystals (SB p.203)
Q.13 (b) CaF2
Only alternate cubical sites are occupied by Ca2+ in order to maintain electroneutrality.
163
Structure of Sodium oxide Structure of Sodium oxide Link Link
7.4 Ionic Crystals (SB p.203)
Na2O vs CaF2
Antifluorite structure
fluorite structure
164
Q.14(a)
Zinc blende structure - Link
165
Q.14(a)
Number of Zn2+ = 4
Number of S2 = 42
16
8
18
ZnS
166
414.0402.0184.0
074.0
r
r
tetrahedral site
S2 ions adopt f.c.c. structure
Alternate Td sites are occupied by Zn2+ ions to ensure electroneutrality.
Zinc sulphide, ZnS
Q.14(b)
167
Q.15
The unit cell is not a cube
Rutile structure - Link
168
Q.15(a)
Number of O2 = 42
142
TiO2
Ti4+O2
Number of Ti4+ = 28
181
169
Q.15(a)
C.N. of Ti4+ = 6
C.N. of O2 = 3
6 : 3 coordination
170
414.0486.0140.0
068.0
r
r
octahedral site
O2 ions adopt distorted h.c.p.(not f.c.c.) structure with alternate Oh sites occupied by Ti4+ ions to ensure electroneutrality.
Titanium(IV) oxide, TiO2
Q.15(b)
171
a
b
a
hexagonal close-packed
172
Factors governing the structures of ionic crystals
1. Close Packing Considerations
Ions in ionic crystals tend to pack as closely as possible. (Why ?)
To strengthen the ionic bonds formed
To maximize the no. of ionic bonds formed.
173
Factors governing the structures of ionic crystals
1. Close Packing Considerations
The larger anions tend to adopt face-centred cubic structure (cubic closest packed, c.c.p.)
The smaller cations tend to fill the interstitial sites as efficiently as possible
174
Factors governing the structures of ionic crystals
Q.16 Is there no limit for the C.N. ? Explain your answer.
The anions tend to repel one another when they approach a given cation.
The balance between attractive forces and repulsive forces among ions limits the C.N. of the system.
175
Factors governing the structures of ionic crystals
Q.16 Is there no limit for the C.N. ? Explain your answer.
If the cations are small, less anions can approach them without significant repulsions.
Or, if the cations are small, they choose to fit in the smaller tetrahedral sites with smaller C.N. to strengthen the ionic bonds formed.
176
Factors governing the structures of ionic crystals
Q.16 Is there no limit for the C.N. ? Explain your answer.
If the cations are large, they choose to fit in the larger octahedral sites or even cubical sites with greater C.N. to maximize the no. of ionic bonds formed.
177
Factors governing the structures of ionic crystals
r
r2. The relative size of cation and anion,
In general, r > r+,
10
r
r
..NCr
r
178
Factors governing the structures of ionic crystalsIf the cations are large, 1
r
r
more anions can approach the cations without significant repulsions.
Or, the large cations can fit in the larger interstitial sites with greater C.N. to maximize the no. of bonds formed.
179
Interstitial site
Tetrahedral Octahedral Cubical
Coordination 4 : 4 6 : 6 8 : 8
0.225 – 0.414 0.414 – 0.732 0.732 – 1.000
ExamplesZnS, most copper(I) halides
Alkali metal halides except C
sCl
CsCl, CsBr
CsI, NH4Cl
7.4 Ionic Crystals (SB p.203)
Summary : -Summary : -
r
r
NH4+ is large
180
7.4 Ionic Crystals (SB p.203)
Summary : -Summary : -
r
r
*CaF2, BaF2, BaCl2, SrF2Examples
0.732 – 1.000
8 : 4Coordinatio
n
CubicalInterstitial
site
181
Q.17
A B
C
r
r+
r
45coscos
rr
r
BC
BAABC
414.0
r
r
45°
182
414.0
r
r
the range of ratio that favours octahedral arrangement.
414.0732.0
r
r
183
732.0
r
r
the optimal ratio for cations and anions to be in direct contact with each other in the cubical sites.
Q.18
184
414.0696.0181.0
126.0
r
r
octahedral site
Cl ions adopt f.c.c. structure
All Oh sites are occupied by Ag+ ions to ensure electroneutrality.
Silver chloride, AgCl
Q.18(a)
185
414.0696.0181.0
126.0
r
r
octahedral site
Ag+ ions adopt an open f.c.c. structure
All Oh sites are occupied by Cl ions to ensure electroneutrality.
Silver chloride, AgCl
Q.18(a)
186
Copper(I) bromide, CuBr
414.0379.0195.0
074.0
r
r
tetrahedral site
Br ions adopt f.c.c. structure
Only alternate Td sites are occupied by Cu+ ions to ensure electroneutrality.
Q.18(b)
187
Copper(II) bromide, CuBr2
414.0364.0195.0
071.0
r
r
tetrahedral site
Br ions adopt f.c.c. structure
Only 1/4 Td sites are occupied by Cu+ ions to ensure electroneutrality.
Some Br may form less bonds than others
Not favourable. Q.18(c)
188
Copper(II) bromide, CuBr2
All the Td sites are occupied by Br ions to ensure electroneutrality.
Q.18(c)
Or, Cu2+ ions adopt a very open f.c.c. structure
However, this structure is too open to form strong ionic bonds.
non-cubic structure is preferred.
189
Q.18(c)
4 : 2 coordinationCu2+
Br
190
Copper(II) chloride, CuCl2
Layer structure
0.295nm0.230 nm
191
Copper(II) chloride, CuCl2
6 : 3 coordination
192
7.7.55 Ionic RadiiIonic Radii
193
Ionic RadiiIonic RadiiIonic radius is the approximate radius of the spherical space occupied by the electron cloud of an ion in all directions in the ionic crystal.
194
Q.19
Why is the electron cloud of an ion always spherical in shape ?
Stable ions always have fully-filled electron shells or subshells. The symmetrical distribution of electrons accounts for the spherical shapes of ions.
195
Q.19
Li+ 1s2
Na+ [He] 2s2, 2px2, 2py
2, 2pz2
Cl [Ne] 3s2, 3px2, 3py
2, 3pz2
Zn2+ [Ar] 3dxy2, 3dxz
2, 3dyz2, ,
222 yx
3d
22z
3d
spherical Symmetrical distribution along x, y and z axes almost spherical
Symmetrical distribution along xy, xz, yz planes and along x, y and z axes almost spherical
196
Determination of Ionic Radii Pauling Scale
Interionic distance (r+ + r) can be determined by
X-ray diffraction crystallography
197
7.5 Ionic Radii (SB p.205)
Electron density plot for sodium chloride crystal
Cl
Cl
Contour having same electron density
Na+
Na+
r++ r
Electron density map for NaCl
r++ r
198
By additivity rule,
Interionic distance = r+ + r
For K+Cl,
r+ + r = 0.314 nm (determined by X-ray)
Since K+(2,8,8) and Cl(2,8,8) are isoelectronic, their ionic radii can be calculated.
r+(K+) = 0.133 nm, r(Cl) = 0.181 nm
199
For Na+Cl,
r+ + r = 0.275 nm (determined by X-ray)
Since r(Cl) = 0.181 nm(calculated)
r+(Na+) = (0.275 - 0.181) nm = 0.094 nm
200
Limitation of Additivity rule
In vacuum, the size of a single ion has no limit according to quantum mechanics.
The size of the ion is restricted by other ions in the crystal lattice.
Interionic distance < r+ + r
201
Evidence :Electron distribution is not perfectly spherical at the boundary due to repulsion between electron clouds of neighbouring ions.
202
Ionic radius depends on the bonding environment.
For example, the ionic radius of Cl ions in NaCl (6 : 6 coordination) is different from that in CsCl (8 : 8 coordination).
203
Ionic radius vs atomic radiusIonic radius vs atomic radius
7.5 Ionic Radii (SB p.206)
Radii of cations < Radii of corresponding parent atoms cations have one less electron shell than the parent
atoms
204
Ionic radius vs atomic radiusIonic radius vs atomic radius
7.5 Ionic Radii (SB p.206)
p/e of cation > p/e of parent atom Less shielding effect stronger nuclear attraction for outermost electrons
smaller size
205
Ionic radius vs atomic radiusIonic radius vs atomic radius
7.5 Ionic Radii (SB p.206)
Radii of anions > Radii of corresponding parent atoms
206
Ionic radius vs atomic radiusIonic radius vs atomic radius
7.5 Ionic Radii (SB p.206)
p/e of anion < p/e of parent atom more shielding effect
weaker nuclear attraction for outermost electrons
larger size
207
Periodic trends of ionic radius
1. Ionic radius increases down a Group
Ionic radius depends on the size of the outermost electron cloud.
On moving down a Group, the size of the outermost electron cloud increases as the number of occupied electron shells increases.
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Periodic trends of ionic radius
2. Ionic radius decreases along a series of isoelectronic ions of increasing nuclear charge
The total shielding effects of isoelectronic ions are approximately the same.
Zeff nuclear charge (Z)
Ionic radius decreases as nuclear charge increases.
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7.5 Ionic Radii (SB p.206)
Isoelectronic to He(2)
Isoelectronic to Ne(2,8)
Isoelectronic to Ar(2,8,8)
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7.5 Ionic Radii (SB p.206)
H is larger than most ions, why ?
211
Q.20 H > N3
The nuclear charge (+1) of H is too small to hold the two electrons which repel each other strongly within the small 1s orbital.
Or,
p/e of H (1/2) < p/e of N3 (7/10)
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The END
Check Point 7-5Check Point 7-5Example 7-5Example 7-5
213
Why do two atoms bond together?
The two atoms tend to achieve an octet configuration which brings stability.
Answer
Introduction (SB p.186)
214
How does covalent bond strength compare with ionic bond strength?
Back
They are similar in strength.
Both are electrostatic attractions between charged particles.
Answer
Introduction (SB p.186)
215
Given the following data:
ΔH (kJ mol–1)
First electron affinity of oxygen –142
Second electron affinity of oxygen +844
Standard enthalpy change of atomization of oxygen +248
Standard enthalpy change of atomization ofaluminium +314
Standard enthalpy change of formation ofaluminium oxide –1669
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
216
ΔH (kJ mol–1)
First ionization enthalpy of aluminium +577
Second ionization enthalpy of aluminium +1820
Third ionization enthalpy of aluminium +2740
(a) (i) Construct a labelled Born-Haber cycle for the formation of aluminium oxide.
(Hint: Assume that aluminium oxide is a purely ionic compound.)
(ii) State the law in which the enthalpy cycle in (i) is based on.
(b) Calculate the lattice enthalpy of aluminium oxide.
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
Answer
217
7.2 Energetics of Formation of Ionic Compounds (SB p.195)
(a) (i)
(ii) The enthalpy cycle in (i) is based on Hess’s law which states that the total enthalpy change accompanying a chemical reaction is independent of the route by means of which the chemical
reaction is brought about.
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7.2 Energetics of Formation of Ionic Compounds (SB p.195)
(b) ΔHf [Al2O3(s)] = 2 × ΔHatom[Al(s)] + 2 × (ΔHI.E.1 [Al(g)]
+ ΔHI.E.2 [Al(g)] + ΔHI.E.3 [Al(g)])
+ 3 × ΔHatom [O2(g)] + 3 × (ΔHE.A.1 [O(g)]
+ ΔHE.A.2 [O(g)]) + ΔHlattice[Al2O3(s)]
ΔHf [Al2O3(s)] = 2 × (+314) + 2 × (+577 + 1 820 + 2 740)
+ 3 × (+248) + 3 × (–142 + 844)
+ ΔHlattice [Al2O3(s)]
ΔHf [Al2O3(s)] = + 628 + 10 274 + 744 + 2 106 + ΔHlattice[Al2O3(s)]
ΔHlattice[Al2O3(s)] = ΔHf [Al2O3(s)] – (628 + 10 274 + 744 + 2 106)
= –1 669 – (628 + 10 274 + 744 + 2 106)
= –15 421 kJ mol–1
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219
(a) Draw a Born-Haber cycle for the formation of magnesium oxide.
(a) The Born-Haber cycle for the formation of MgO:
Answer
7.2 Energetics of Formation of Ionic Compounds (SB p.197)
220
(b)Calculate the lattice enthalpy of magnesium oxide by means of the Born-Haber cycle drawn in (a).
Given: ΔHatom [Mg(s)] = +150 kJ mol–1
ΔHI.E. [Mg(g)] = +736 kJ mol–1
ΔHI.E. [Mg+(g)] = +1 450 kJ mol–1
ΔHatom [O2(g)] = +248 kJ mol–1
ΔHE.A. [O(g)] = –142 kJ mol–1
ΔHE.A. [O–(g)] = +844 kJ mol–1
ΔHf [MgO(s)] = –602 kJ mol–1
7.2 Energetics of Formation of Ionic Compounds (SB p.197)
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Answer
221
Back
7.2 Energetics of Formation of Ionic Compounds (SB p.197)
(b) ΔHlattice [MgO(s)]
= ΔHf [MgO(s)] – ΔHatom [Mg(s)]
– ΔHI.E. [Mg(g)] – ΔHI.E. [Mg+(g)] – ΔHatom [O2(g)]
– ΔHE.A. [O(g)] – ΔHE.A. [O–(g)]
= [–602 – 150 – 736 – 1 450 – 248 –(–142) – 844] kJ mol–1
= –3 888 kJ mol–1
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222
Give two properties of ions that will affect the value of the lattice enthalpy of an ionic compound.The charges and sizes of ions will affect the value of the lattice enthalpy.
The smaller the sizes and the higher the charges of ions, the higher (i.e. more negative) is the lattice enthalpy.
Answer
Back
7.3 Stoichiometry of Ionic Compounds (SB p.201)
223
Write down the formula of the compound that possesses the lattice structure shown on the right:
To calculate the number of each type of particle present in the unit cell:
Number of atom A = 1
(1 at the centre of the unit cell)
Number of atom B = 8 × = 2
(shared along each edge)
Number of atom C = 8 × = 1
(shared at each corner)
∴ The formula of the compound is AB2C.
41
41
81 8
1
Answer
Back7.4 Ionic Crystals (SB p.204)
224
The diagram on the right shows a unit cell of titanium oxide. What is the coordination number of
(a) titanium; and
(b) oxygen?(a) The coordination number of titanium is 6 as there are six oxide ions surrounding each titanium ion.
(b) The coordination number of oxygen is 3.
Answer
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7.4 Ionic Crystals (SB p.205)
225
The following table gives the atomic and ionic radii of some Group IIA elements.
7.5 Ionic Radii (SB p.208)
Element Atomic radius (nm) Ionic radius
Be 0.112 0.031
Mg 0.160 0.065
Ca 0.190 0.099
Sr 0.215 0.133
Ba 0.217 0.135
226
Explain briefly the following:
(a) The ionic radius is smaller than the atomic radius in each element.
(b) The ratio of ionic radius to atomic radius of Be is the lowest.
(c) The ionic radius of Ca is smaller than that of K(0.133 nm).
7.5 Ionic Radii (SB p.208)
Answer
227
7.5 Ionic Radii (SB p.208)
(a) One reason is that the cation has one electron shell less than the corresponding atom. Another reason is that in the cation, the number of protons is greater than the number of electrons. The electron cloud of the cation therefore experiences a greater nuclear attraction. Hence, the ionic radius is smaller than the atomic radius in each element.
(b) In the other cations, although there are more protons in the nucleus, the outer most shell electrons are further away from the nucleus, and electrons in the inner shells exhibit a screening effect. Be has the smallest atomic size. In Be2+ ion, the electrons experience the greatest nuclear attraction. Therefore, the contraction in size of the electron cloud is the greatest when Be2+ ion is formed, and the ratio of ionic radius to atomic radius of Be is the lowest.
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7.5 Ionic Radii (SB p.208)
(c) The electronic configurations of both K+ and Ca2+ ions are 1s22s22p63s23p6. Hence they have the same number and arrangement of electrons. However, Ca2+ ion is doubly charged while K+ ion is singly charged, so the outermost shell electrons of Ca2+ ion experience a greater nuclear attraction. Hence, the ionic radius of Ca2+ ion is smaller than that of K+ ion.
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229
Arrange the following atoms or ions in an ascending order of their sizes:
(a) Be, Ca, Sr, Ba, Ra, Mg
(b) Si, Ge, Sn, Pb, C
(c) F–, Cl–, Br–, I–
(d) Mg2+, Na+, Al3+, O2–, F–, N3–
(a) Be < Mg < Ca < Sr < Ba < Ra
(b) C < Si < Ge < Sn < Pb
(c) F– < Cl– < Br– < I–
(d) Al3+ < Mg2+ < Na+ < F– < O2– < N3–
Answer
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7.5 Ionic Radii (SB p.208)